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The iterated Cauchy integral formula on a polydisc

Statement

Fix m1, a point aCm and a polyradius ρ. Let f:Δρ(a)C be continuous and separately holomorphic (Separately holomorphic functions), let r be a polyradius with rk<ρk for every k<m, and let Ck(t)=ak+rkexp(it) on [0,2π]. Then for every zΔr(a)

f(z)=1(2πi)mC0 ⁣ ⁣Cm1f(ζ0,,ζm1)k<m(ζkzk)dζm1dζ0.

The right-hand side is an iterated integral: the innermost integral is taken over ζm1 with ζ0,,ζm2 held fixed, then over ζm2, and so on. Each successive integrand is continuous on the circle it is integrated over, so each of the m integrals exists. No integral over the distinguished boundary is formed and the order of integration is never interchanged.

Facts & Assumptions

Given: m1, aCm, polyradii ρ and r with rk<ρk, a continuous separately holomorphic f:Δρ(a)C, the circles Ck, and zΔr(a); Cm is read through Complex m-space and its real coordinate dictionary.

[L1]

Δr(a), Δr(a) and Γr(a) are defined coordinatewise by zkak<rk, rk and =rk (Balls, polydiscs and the distinguished boundary in Cm).

[L2]

f is separately holomorphic when for every bU and k<m the slice ζf(b0,,bk1,ζ,bk+1,,bm1) is holomorphic on the open set of ζ for which the point lies in U (Separately holomorphic functions).

[L3]

If f is holomorphic on D(a,R), 0<r<R, za<r and γ(t)=a+rexp(it) on [0,2π], then f(z)=(2πi)1γf(ζ)(ζz)1dζ (Cauchy's integral formula on a circle compactly contained in a disc of holomorphy).

[L5]

If a property holds at 0 and passes from q to q+1, it holds for every natural number (The principle of mathematical induction).

[L6]

For aC, r>0 and kZ, the contour a+rexp(ikt) on [0,2π] is a closed complex contour whose trace for k0 is {a=r} (A circle traversed k times has winding number k inside and 0 outside).

[L7]

zw=zw and z+wz+w (Conjugation is an involutive real-field automorphism, zz=z2, and modulus is definite, multiplicative, and subadditive).

[L8]

Nonvanishing quotients of functions complex differentiable at a point are complex differentiable there (Linearity, product, reciprocal, and quotient rules for complex derivatives), such functions are continuous (Complex differentiability at a point implies continuity there), and composites of continuous maps are continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

Proof

technique · direct
1.1

By [L6] each Ck is a closed complex contour with trace the circle {ζak=rk}. If ζjaj=rj for j<p and zjaj<rj for jp, then ζjaj<ρj and zjaj<ρj by the hypothesis rj<ρj, so every such mixed point lies in Δρ(a) by [L1].

givenL1L6
1.2

For the fixed point z, define Hm(ζ0,,ζm1;z):=f(ζ0,,ζm1). Then, for p=m1,m2,,0, define Hp(ζ0,,ζp1;z) by Hp:=12πiCpHp+1(ζ0,,ζp1,ζp;z)ζpzpdζp, whenever the integrand is continuous on Cp. By construction H0(;z) is exactly the iterated integral in the statement, divided by (2πi)m.

given
2.1

Claim, proved by induction on q=mp using [L5]: for every p with 0pm the quantity Hp is defined and Hp(ζ0,,ζp1;z)=f(ζ0,,ζp1,zp,,zm1). For q=0, that is p=m, this is the definition of Hm.

step 1.1step 1.2L5
2.2

The slice ξf(ζ0,,ζp1,ξ,zp+1,,zm1) is holomorphic on the disc ξap<ρp: by step 1.1 the corresponding point lies in Δρ(a) for every such ξ, and by [L1] and [L2] that disc is exactly the slice domain, on which separate holomorphy makes the slice holomorphic.

step 1.1L1L2
3.1

Assume the claim for p+1. Fix ζ0,,ζp1 on their circles. By the assumption, Hp+1(ζ0,,ζp1,ζp;z)=f(ζ0,,ζp,zp+1,,zm1), which by step 1.1 is a continuous function of ζp on the circle ζpap=rp; dividing by ζpzp, which is nonzero there because zpap<rp by [L1] and [L7], leaves a continuous integrand by [L8], so the integral defining Hp exists by [L4].

step 1.1step 2.1L1L4L7L8
4.1

Applying [L3] to the slice of step 2.2, with R=ρp, r=rp and z=zp, gives f(ζ0,,ζp1,zp,zp+1,,zm1)=12πiCpf(ζ0,,ζp1,ζp,zp+1,)ζpzpdζp, which by step 3.1 is Hp(ζ0,,ζp1;z). This is the claim for p, so the induction of step 2.1 closes.

step 3.1step 2.2L3
5.1

Taking p=0 in step 2.1 gives H0(;z)=f(z), and step 1.2 identifies H0(;z) with the iterated integral divided by (2πi)m; every one of the m integrals exists by step 3.1. Since zΔr(a) was arbitrary, the formula holds throughout Δr(a).

step 1.2step 2.1step 3.1step 4.1

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